Petroleum Reserves
Estimation Methods
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Material Balance Calculations for Oil Reservoirs
A general material balance equation that can be applied to all reservoir types was
first developed by Schilthuis in 1936. Although it is a tank model equation, it can
provide great insight for the practicing reservoir engineer. It is written from start of
production to any time (t) as follows:
Expansion of oil in the oil zone +
Expansion of gas in the gas zone +
Expansion of connate water in the oil and gas zones +
Contraction of pore volume in the oil and gas zones +
Water influx + Water injected + Gas injected =
Oil produced + Gas produced + Water produced
Mathematically, this can be written as:
C S 
N ( B t - B ti ) + G ( B g - B gi ) + ( NB ti + GB gi )  w wi  ∆ p t
 1 - S wi 
 Cf 
+ ( NB ti + GB gi ) 
 ∆ p t + W e + W I B Iw + G I B Ig
 1 - S wi 
= N p B t + N p ( R p - R soi ) B g + W p B w
Where:
N
Np
G
GI
Gp
We
WI
Wp
Bti
Boi
Bgi
Bt
Bo
Bg
Bw
BIg
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
initial oil in place, STB
cumulative oil produced, STB
initial gas in place, SCF
cumulative gas injected into reservoir, SCF
cumulative gas produced, SCF
water influx into reservoir, bbl
cumulative water injected into reservoir, STB
cumulative water produced, STB
initial two-phase formation volume factor, bbl/STB = Boi
initial oil formation volume factor, bbl/STB
initial gas formation volume factor, bbl/SCF
two-phase formation volume factor, bbl/STB = Bo + (Rsoi - Rso) Bg
oil formation volume factor, bbl/STB
gas formation volume factor, bbl/SCF
water formation volume factor, bbl/STB
injected gas formation volume factor, bbl/SCF
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BIw
Rsoi
Rso
Rp
Cf
Cw
Swi
∆pt
p(t)
=
=
=
=
=
=
=
=
=
injected water formation volume factor, bbl/STB
initial solution gas-oil ratio, SCF/STB
solution gas-oil ratio, SCF/STB
cumulative produced gas-oil ratio, SCF/STB
formation compressibility, psia-1
water isothermal compressibility, psia-1
initial water saturation
reservoir pressure drop, psia = pi - p(t)
current reservoir pressure, psia
The MBE as a Straight Line
Normally, when using the material balance equation, each pressure and the
corresponding production data is considered as being a separate point from
other pressure values. From each separate point, a calculation is made and the
results of these calculations are averaged. However, a method is required to
make use of all data points with the requirement that these points must yield
solutions to the material balance equation that behave linearly to obtain values
of the independent variable. The straight-line method begins with the material
balance written as:
 C f + C w S wi 
N ( B t - B ti ) + G ( B g - B gi ) + ( NB ti + GB gi ) 
 ∆ pt
 1 - S wi 
+ W e + W I B Iw + G I B Ig
= N p B t + N p ( R p - R soi ) B g + W p B w
Defining the ratio of the initial gas cap volume to the initial oil volume as:
m=
initial gas cap volume GBgi
=
NBti
initial oil volume
and plugging into the equation yields:
N ( B t - B ti ) + Nm
+


B ti
( B g - B gi ) + ( NB ti + Nm B ti )  C f1 - C w S wi  ∆ p t
S wi 
B gi

+ W e + W I B Iw + G I B Ig
= N p B t + N p ( R p - R soi ) B g + W p B w
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Let:
E o = B t - B ti
Eg=
E
F=N
p
f,w
 Bt +

B ti
( B g - B gi )
B gi
 C f + C w S wi 
= (1 + m ) B ti 
 ∆ pt
1 - S wi


( R p - R soi ) B g
 +W

p
Bw -W
I
B Iw - G I B Ig
Thus we obtain:
F = NEo + mNE g + NE f ,w + W e
= N ( Eo + mE g + E f ,w ) + We
The following cases are considered:
1.
No gas cap, negligible compressibilities, and no water
influx
F= NEo
2.
Negligible compressibilities, and no water influx
F= NEo + NmE g
Eg
F
= N + Nm
Eo
Eo
Which is written as y = b + x. This would suggest that a plot of F/Eo as
the y coordinate versus Eg/Eo as the x coordinate would yield a straight
line with slope equal to mN and intercept equal to N.
3.
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Including compressibilities and water influx, let:
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Estimation Methods
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D = Eo + mE g + E f ,w
Dividing through by D, we get:
F
W
=N+ e
D
D
Which is written as y = b + x. This would suggest that a plot of F/D as
the y coordinate and We/D as the x coordinate would yield a straight line
with slope equal to 1 and intercept equal to N.
Drive Indexes from the MBE
The three major driving mechanisms are:
1. Depletion drive (oil zone oil expansion),
2. Segregation drive (gas zone gas expansion), and
3. Water drive (water zone water influx).
To determine the relative magnitude of each of these driving mechanisms, the
compressibility term in the material balance equation is neglected and the
equation is rearranged as follows:
N ( B t - B ti ) + G ( B g - B gi ) + (W e - W p B w ) = N p  B t + ( R p - R soi ) B g 
Dividing through by the right hand side of the equation yields:
N ( B t - B ti )
N p  B t + ( R p - R soi ) B g 
+
G ( B g - B gi )
N p  B t + ( R p - R soi ) B g 
+
(W e - W p B w )
=1
N p  B t + ( R p - R soi ) B g 
The terms on the left hand side of equation (3) represent the depletion drive
index (DDI), the segregation drive index (SDI), and the water drive index
(WDI) respectively. Thus, using Pirson's abbreviations, we write:
DDI + SDI + WDI = 1
The following examples should clarify the errors that creep in during the
calculations of oil and gas reserves.
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Estimation Methods
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Example #5: Given the following data for an oil field
Volume of bulk oil zone
Volume of bulk gas zone
Initial reservoir pressure
Initial oil FVF
Initial gas FVF
Initial dissolved GOR
Oil produced during the interval
Reservoir pressure at the end of the interval
Average produced GOR
Two-phase FVF at 2000 psia
Volume of water encroached
Volume of water produced
Water FVF
Gas FVF at 2000 psia
=
=
=
=
=
=
=
=
=
=
=
=
=
=
112,000 acre-ft
19,600 acre-ft
2710 psia
1.340 bbl/STB
0.006266 ft3/SCF
562 SCF/STB
20 MM STB
2000 psia
700 SCF/STB
1.4954 bbl/STB
11.58 MM bbl
1.05 MM STB
1.028 bbl/STB
0.008479 ft3/SCF
The following values will be calculated:
1. The stock tank oil initially in place.
2. The driving indexes.
3. Discussion of results.
Solution:
1. The material balance equation is written as:
N (Bt - Bti ) + G (B g - B gi ) = N p [ Bt + (R p - R soi ) B g ] - (W e - W p B w )
Define the ratio of the initial gas cap volume to the initial oil volume as:
m=
GB gi
NBti
we get:
N (Bt - Bti ) + Nm Bti (B g - B gi ) = N p [ Bt + (R p - R soi ) B g ] - (W e - W p B w )
B gi
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Estimation Methods
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and solve for N, we get:
N=
N p [ Bt + (R p - R soi ) B g ] - (W e - W p B w )
(Bt - Bti ) + m Bti (B g - B gi )
B gi
Since:
Np
Bt
Rp
Rsoi=
Bg
We
Wp
Bw
Bti
m
Bgi
= 20 x 106 STB
= 1.4954 bbl/STB
= 700 SCF/STB
562 SCF/STB
= 0.008479 ft3/SCF = 0.008479/5.6146 = 0.001510 bbl/SCF
= 11.58 x 10 6 bbl
= 1.05 x 106 STB
= 1.028 bbl/STB
= 1.34 bbl/STB
= GBgi/NBti = 19,600/112,000 = 0.175
= 0.006266 ft3/SCF = 0.006266/5.6146 = 0.001116 bbl/SCF
Thus:
N=
20  1.4954 + ( 700 - 562 ) 0.001510  - ( 11.58 - 1.05x1.028 ) 6
10
1.34
(1.4954 - 1.34 ) + 0.175
(0.001510 - 0.001116 )
0.001116
= 98.97 MM STB
2. In terms of drive indexes, the material balance equation is written as:
N ( B t - B ti
)
N p  B t + ( R p - R soi ) B g 
+
G ( B g - B gi )
N p  B t + ( R p - R soi ) B g 
+
( W e - W p Bw )
=1
N p  B t + ( R p - R soi ) B g 
Thus the depletion drive index (DDI) is given by:
N ( B t - B ti )
N p  B t + ( R p - R soi ) B g 
=
98.97x10 6 ( 1.4954 - 1.34 )
20x10 6 1.4954 + (700 - 562 ) 0.001510 
= 0.45
The segregation drive index (SDI) is given by:
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Nm B ti ( B g - B gi )
B gi
=

N p  B t + ( R p - R soi ) B g 
1.34
( 0.001510 - 0.001116 )
0.001116
= 0.24
20x10 6 1.4954 + (700 - 562 ) 0.001510 
98.97 x 10 6 x 0.175x
The water drive index (WDI) is given by:
( W e - W p Bw ) = ( 11.58 x 106 - 1.05 x 1.028 x 106 ) = 0.31
6
N p [ Bt + (R p - R soi ) B g ] 20x 10 [1.4954 + (700 - 562 ) 0.001510 ]
3. The drive mechanisms as calculated in part (2) indicate that when the
reservoir pressure has declined from 2710 psia to 2000 psia, 45% of the
total production was by oil expansion, 31% was by water drive, and 24%
was by gas cap expansion.
This concludes the solution.
Example #6: Given the following data for an oil field
A gas cap reservoir is estimated, from volumetric calculations, to have an
initial oil volume N of 115 x 106 STB. The cumulative oil production Np and
cumulative gas oil ratio Rp are listed in the following table as functions of the
average reservoir pressure over the first few years of production. Assume
that pi = pb = 3330 psia. The size of the gas cap is uncertain with the best
estimate, based on geological information, giving the value of m = 0.4. Is this
figure confirmed by the production and pressure history? If not, what is the
correct value of m?
Pressure
psia
3330
3150
3000
2850
2700
2550
2400
Np
MM STB
0
3.295
5.903
8.852
11.503
14.513
17.73
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Rp
SCF/STB
0
1050
1060
1160
1235
1265
1300
Bo
BBL/STB
1.2511
1.2353
1.2222
1.2122
1.2022
1.1922
1.1822
Rso
SCF/STB
510
477
450
425
401
375
352
Bg
bbl/SCF
0.00087
0.00092
0.00096
0.00101
0.00107
0.00113
0.00120
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Petroleum Reserves
Estimation Methods
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Solution:
Calculate the parameters F, Eo, Eg as given by the above equations:
Bt
BBL/STB
1.2511
1.26566
1.2798
1.29805
1.31883
1.34475
1.3718
F
MM/RB
Eo
RB/STB
Eg
RB/SCF
F/Eo
MM/STB
Eg/Eo
5.8073
10.6714
17.3017
24.0940
31.8981
41.1301
0.014560
0.028700
0.046950
0.067730
0.093650
0.120700
0.071902299
0.129424138
0.201326437
0.287609195
0.373891954
0.474555172
398.8534135
371.8272962
368.5128136
355.7353276
340.6099594
340.7626678
4.938344701
4.509551844
4.28810302
4.246407728
3.992439445
3.931691569
The plot of F/Eo versus Eg/Eo is shown next:
Chart Title
420
400
F/Eo
380
360
340
y = 58.83x + 108.7
320
300
3.8
4
4.2
4.4
4.6
4.8
5
5.2
Eg/Eo
Figure 6: F/Eo vs. Eg/Eo Plot
The best fit is expressed by:
Eg
F
= 108.7 + 58.83
Eo
Eo
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Therefore, N = 108.7 MM STB and m = 58.83/108.7 = 0.54.
This concludes the solution of this problem.
Example #7: Given the following data for an oil field
A gas cap reservoir is estimated, from volumetric calculations, to have an
initial oil volume N of 47 x 106 STB. The cumulative oil production Np and
cumulative gas oil ratio Rp are listed in the following table as functions of the
average reservoir pressure over the first few years of production. Other
pertinent data are also supplied. Assume pi = pb = 3640 psia. The size of the
gas cap is uncertain with the best estimate, based on geological information,
giving the value of m = 0.0. Is this figure confirmed by the production
history? If not, what is the correct value of m?
Pressure
psia
3640
3585
3530
3460
3385
3300
3200
Bg
bbl/SCF
0.000892
0.000905
0.000918
0.000936
0.000957
0.000982
0.001014
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pi =
Cf =
Cw =
Swi =
Bw =
m=
3640
0.000004
0.000003
0.25
1.025
0
psia
psia-1
psia-1
Np
MM STB
0
0.79
1.21
1.54
2.08
2.58
3.4
Gp
MM SCF
0
4.12
5.68
7
8.41
9.71
11.62
Bt
BBL/STB
1.464
1.469
1.476
1.482
1.491
1.501
1.519
We
MM BBL
0
48.81
61.187
71.32
80.293
87.564
93.211
Wp
MM STB
0
0.08
0.26
0.41
0.6
0.92
1.38
psia
Rp
SCF/STB
0
5.215189873
4.694214876
4.545454545
4.043269231
3.763565891
3.417647059
Rso
SCF/STB
888
874
860
846
825
804
779
F
MM/RB
0.6114
1.0713
1.4291
1.9567
2.5753
3.5294
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Solution:
Calculate the parameters F, Eo, Eg, Ef,w, and D, as given by the above
equations:
Eo
RB/STB
Eg
RB/SCF
Ef,w
RB/SCF
D
F/D
MM/STB
We/D
0.005000
0.012000
0.018000
0.027000
0.037000
0.055000
0.021336323
0.042672646
0.072215247
0.106681614
0.147713004
0.200233184
0.00050996
0.00101992
0.00166896
0.00236436
0.00315248
0.00407968
0.00550996
0.01301992
0.01966896
0.02936436
0.04015248
0.05907968
110.9559779
82.28173445
72.65677901
66.63557762
64.1383531
59.739895
8858.50351
4699.491241
3626.017847
2734.369147
2180.786841
1577.716738
The plot of F/D versus We/D is shown next. The best fit is expressed by:
W
F
= 0.0071 + 48.067e6 e
D
D
Therefore, N = 48 MM STB and m = 0.0071.
This concludes the solution of this problem.
Chart Title
119
109
F/Eo
99
89
79
69
y = 0.0071x + 48.067
59
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
Eg/Eo
Figure 7: F/Eo vs. Eg/Eo Plot
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